Chap19Sol

Chap19Sol - SOLUTIONS TO PROBLEMS Easy(a 10.0 grams \$...

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1 SOLUTIONS TO PROBLEMS Easy P19.2 (a) N = 10.0 grams 107.87 grams mol ! " # \$ % & 6.02 ' 10 23 atoms mol ! " # \$ % & 47 electrons atom ! " # \$ % & = 2.62 ' 10 24 (b) # electrons added = Q e = 1.00 ! 10 " 3 C 1.60 ! 10 " 19 C electron = 6.25 ! 10 15 or 2.38 electrons for every 10 9 already present . P19.4 The force on one proton is r F = k e q 1 q 2 r 2 away from the other proton. Its magnitude is 8.99 ! 10 9 N " m C 2 ( ) 1.6 ! 10 # 19 C 2 ! 10 # 15 m \$ % & ' ( ) 2 = 57.5 N . P19.7 (a) The force is one of attraction . The distance r in Coulomb’s law is the distance between centers. The magnitude of the force is F = k e q 1 q 2 r 2 = 8.99 ! 10 9 N " m 2 C 2 ( ) 12.0 ! 10 # 9 C ( ) 18.0 ! 10 # 9 C ( ) 0.300 m ( ) 2 = 2.16 ! 10 # 5 N . (b) The net charge of ! 6.00 " 10 ! 9 C will be equally split between the two spheres, or ! 3.00 " 10 ! 9 C on each. The force is one of repulsion , and its magnitude is F = k e q 1 q 2 r 2 = 8.99 ! 10 9 N " m 2 C 2 ( ) 3.00 ! 10 # 9 C ( ) 3.00 ! 10 # 9 C ( ) 0.300 m ( ) 2 = 8.99 ! 10 # 7 N .

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2 Electric Forces and Electric Fields P19.10 For equilibrium, r F e = ! r F g or q r E = ! mg ! ˆ j ( ) . Thus, r E = mg q ˆ j . (a) r E = mg q ˆ j = 9.11 ! 10 " 31 kg ( ) 2 ( ) " 1.60 ! 10 " 19 C ( ) ˆ j = " 5.58 ! 10 " 11 ( ) ˆ j (b) r E = mg q ˆ j = 1.67 ! 10 " 27 kg ( ) 2 ( ) 1.60 ! 10 " 19 C ( ) ˆ j = 1.02 ! 10 " 7 ( ) ˆ j *P19.24 (a) q 1 q 2 = ! 6 18 = ! 1 3 (b) q 1 is negative, q 2 is positive P19.27 (a) a = qE m = 1.60 ! 10 " 19 640 ( ) 1.67 ! 10 " 27 = 6.13 ! 10 10 2 (b) v f = v i + at 1.20 ! 10 6 = 6.13 ! 10 10 ( ) t t = 1.95 ! 10 " 5 s (c) x f ! x i = 1 2 v i + v f ( ) t x f = 1 2 1.20 ! 10 6 ( ) 1.95 ! 10 " 5 ( ) = 11.7 m (d) K = 1 2 mv 2 = 1 2 1.67 ! 10 " 27 kg ( ) 1.20 ! 10 6 ( ) 2 = 1.20 ! 10 " 15 J P19.30 ! E = EA cos " = 2.00 # 10 4 ( ) 18.0 m 2 ( ) cos10.0 ° = 355 kN \$ m 2 P19.31 ! E = EA cos A = ! r 2 = 0.200 ( ) 2 = 0.126 m 2 5.20 ! 10 5 = E 0.126 ( ) cos0 ° E = 4.14 ! 10 6 N C = 4.14 MN C P19.32 (a) E = k e Q r 2 : 8.90 ! 10 2 = 8.99 ! 10 9 ( ) Q 0.750 ( ) 2 But Q is negative since r E points inward. Q = ! 5.57 " 10 ! 8 C = ! 55.7 nC (b) The negative charge has a spherically symmetric charge distribution, concentric with the spherical shell.